Question

7. A convex lens of focal length 10 cm is placed in contact with a concave lens of focal length 20 The focal length of this combination of lenses will be

(a) + 10 cm

(b) + 20 cm

(c) -10 cm

(d) -20 cm​

Answer 1

Net Focal Length is 20 cm.

Explanation:

As we already know,

$Power (P) = \frac {1}{Focal\ Length} </p><p>Focal Length$

$Hence$

$</p><p>Power of Convex Lens, P_1 = \frac {1}{10 cm}P </p><p>1</p><p> </p><p> = </p><p>10cm$

$= \frac {1}{0.1 m} </p><p>0.1m</p><p>1</p><p> </p><p> = + 10 D</p><p>$

$</p><p>Power of Concave Lens, P_2 = \frac {1}{- 20 cm}P </p><p>2</p><p> </p><p> = </p><p>−20cm</p><p>1= \frac {- 1}{0.2 m} </p><p>0.2m</p><p>−1</p><p> </p><p> = - 5 D$

$Also, When \: lenses \: comes \: into \: contact \: then the \: Len's power \: adds \: up.</h2><h2>So, \: we \: get:$

$Net Power (P) = P_1 + P_2P </p><p>1</p><p> </p><p> +P </p><p>2</p><p> </p><p> </p><p></p><p>Net Power (P) = 10 D - 5 D = 5 D$

$Net Focal Length = \frac {1}{Power} </p><p>Power</p><p>1</p><p> </p><p> </p><p></p><p>= \frac {1}{5} </p><p>5</p><p>1</p><p> </p><p> </p><p></p><p>= 0.2 m</p><p></p><p>= 20 cm$